$f(x)=-\dfrac{81}{\sqrt[{3}]{x+5}}$. On which intervals is the graph of $f$ concave up? Choose 1 answer: Choose 1 answer: (Choice A) A $x<-\sqrt[3]{5}$ only (Choice B) B $x<-5$ only (Choice C) C $x>-5$ only (Choice D) D $x>\sqrt[3]{5}$ only
Solution: We can analyze the intervals where $f$ is concave up/down by looking for the intervals where its second derivative $f''$ is positive/negative. This analysis is very similar to finding increasing/decreasing intervals, only instead of analyzing $f'$, we are analyzing $f''$. The second derivative of $f$ is $f''(x)=-\dfrac{36}{\sqrt[3]{(x+5)^7}}$. $f''$ is never equal to $0$. $f''$ is undefined for $x=-5$. Therefore, our only point of interest is $x=-5$. Our point of interest divides the domain of $f$ (which is all numbers except for $-5$ ) into two intervals: $\llap{-}9$ $\llap{-}8$ $\llap{-}7$ $\llap{-}6$ $\llap{-}5$ $\llap{-}4$ $\llap{-}3$ $\llap{-}2$ $\llap{-}1$ $x< \llap{-}5$ $x> \llap{-}5$ Let's evaluate $f''$ at each interval to see if it's positive or negative on that interval. Interval $x$ -value $f''(x)$ Verdict $x<-5$ $x=-6$ $f''(-6)=36>0$ $f$ is concave up $\cup$ $x>-5$ $x=-4$ $f''(-4)=-36<0$ $f$ is concave down $\cap$ In conclusion, the graph of $f$ is concave up over the interval $x<-5$ only.